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Can you help me with this? I did this many times and got different answers each time.

Find f.
f ''(x) = 3e^x + 3sin(x)
f(0) = 0
f(π) = 0

My work:
f ''(x) = 3e^x + 3sin(t)
f'(x) = 3e^x - 3cos(t) + C
f(x) = 3e^x -3sin(t) + Cx + D

0=f(0)= 3e^0 - 3sin(0) + C(0) + D
D=-3

0=f(π)= 3e^π - 3sin(π) + Cπ -3
C= (3-3e^π)/π

F(x) = 3e^x -3sin(x) + (3-3e^π)/π x - 3

User Icon for MathMate MathMate answered
13 years ago

f(0) = 0

f(π) = 0
Check if one of them is actually f'().

Assuming it's correct as above:
from:
f(x) =3e^x -3sin(x) + (3-3e^π)x/π - 3
We get:
f"(x)=3sin(x)+3e^x
f(0)=0
f(π)=0
which clearly satisfy all the given requirements. However, since there were two initial conditions for f(x), it is possible to have multiple solutions.

User Icon for Explain Bot Explain Bot answered
11 months ago

To find the function f given the second derivative f''(x) and the boundary conditions f(0) = 0 and f(π) = 0, we can follow these steps:

1. Start by integrating the second derivative f''(x) with respect to x to find the first derivative f'(x). Keep in mind that when you integrate, you need to add a constant of integration (C).

∫ (f''(x)) dx = ∫ (3e^x + 3sin(x)) dx
= 3∫ (e^x) dx + 3∫ (sin(x)) dx
= 3(e^x) - 3(cos(x)) + C1

2. Now, integrate the first derivative f'(x) to find the original function f(x). Again, include a constant of integration (D).

∫ (f'(x)) dx = ∫ (3e^x - 3cos(x) + C1) dx
= 3∫ (e^x) dx - 3∫ (cos(x)) dx + ∫ (C1) dx
= 3(e^x) - 3(sin(x)) + C1x + D

3. Apply the boundary conditions f(0) = 0 and f(π) = 0 to determine the values of the constants C1 and D.

For f(0)=0:
f(0) = 3e^0 - 3sin(0) + C1(0) + D
0 = 3 - 0 + 0 + D
D = -3

For f(π)=0:
f(π) = 3e^π - 3sin(π) + C1(π) - 3
0 = 3e^π - 0 + C1π - 3
C1 = (3 - 3e^π)/π

4. Substitute the values of C1 and D back into the equation for f(x) to get the final result:

f(x) = 3e^x - 3sin(x) + ((3 - 3e^π)/π) x - 3